Hello. In this video, we are going to deal with the N-shaped trusses. And we will see how to solve certain cases of particular nodes. We will see that the solving method for trusses with N-shaped diagonals is similar to what we had till now, with simply Cremona diagrams which have a slightly different shape. We will determine the internal forces in each of the bars, and then we will make some observations about these internal forces. Finally, we will spend a bit of time looking at cases of particular nodes, particularly unloaded nodes. In this video, you can see a truss with 10 nodes, which is being to be loaded by three loads of 10N on the three upper nodes, on the middle. Let's first figure out if this structure is statically determinate. Here, we have a fixed support, so one, two support reactions ; a mobile support on the right, we thus have three reaction forces. We are going to count the bars : 1, 2, 3, â€¦ (counts till 17) 17 bars in total. Let's count the nodes : 1, 2, 3,â€¦ (counts till 10) 2 x 10 = 20, which is also equal to 3 + 17 ; therefore this structure is statically determinate. We are going to proceed with the resolution; we already have the support reactions. In this case, it is quite easy since the resultant is located in the middle and has a value of 30 N, the resultant is the applied force. And then, we have a support reaction on the left and on the right, which is equal to 15 N. Let's start by node A, which is slightly particular since a support reaction acts on it, and there is a bar which is lined up with this support reaction ; there is also another bar which is not lined up. Well, for the sum of these three forces to be equal to zero, it is necessary that the force in A-C be equal to zero. Let's build the Cremona diagram, with the vertical reaction R(A) on the right, which is equal to 15 N, and the internal force in the bar A-D which will be equal to 15 N and will act on the opposite way. I draw it slightly staggered, but obviously, in reality, it should be superimposed. So this bar A-B is in compression, and the contribution of the node A to the Cremona diagram is simply this line, or these two superimposed lines. Let's now move to node B which is an unloaded node. This node is subjected to the internal force A-B in the other direction, then, turning counterclockwise, I first meet the bar B-C, then the bar B-D. We can see that the internal force in B-C pulls on the free-body, it is thus a tensile internal force, while the internal force in B-D pushes on the free-body, it is therefore a compressive internal force. This orange triangle indicates the contribution of node B to the Cremona diagram. We now pass to node C, another unloaded node, which is of course subjected to the internal force in A-C, equal to zero, to the internal force in B-C in the other direction, then to the internal force in C-E, horizontally, and finally to the internal force in C-D, vertically. We can see that the internal force in C-E pulls on the free-body, it is thus a tensile internal force, while the internal force in C-D pushes on the free-body, it is therefore a compressive internal force. The contribution of this free-body to the Cremona diagram is thus this pink triangle. We now move to the node D, a loaded node, which is subjected to a load of 10N ; I am going to draw it, one more time, in a staggered way ; while in reality, it should be superimposed. Here, I have a load of 10 N ; then, turning counterclockwise, I meet the internal force in B-D in the other direction, then the internal force in C-D in the other direction, now I meet the bar D-E, then, I meet the bar D-F, to finally come back to the beginning of the load of 10 N. The internal force in D-E pulls on the free-body, it is a tensile internal force. And the internal force in D-F pushes on the free-body, it is D-F, here ; the contribution of this node D to the global equilibrium is thus like this. We cannot move now to the node E, since there are still one, two, three bars in which the internal forces are unknown, but we can move to the node F. Again, this node is subjected to a load of 10 N. Then, the internal force in D-F in the other direction, the internal force in F-E, and finally the internal force in F-H. The internal force in E-F pushes on the free-body, so let's say E-F, it is then a compressive internal force, such as the internal force in F-H. The contribution of the yellow node to the Cremona diagram is this rectangle, with the same internal forces F-H and D-F, and 10N for the internal force in the bar E-F. We can now look at the equilibrium of the node E. We know the internal force in E-F, well, however, in the other direction ; the internal force in D-E, in the other direction, the internal force in C-E, in the other direction, then, we come back by an internal force which is going to be equal and in the opposite direction, -- I am not going to redraw it, butâ€¦ this is the internal force in E-G. We could have guessed that, since the structure is symmetrical, the internal forces must also obviously be symmetrical. Then, by an internal force in E-H, which will also be symmetrical to the internal force in D-E, â€¦ E-H, here. This light blue node contributes to the Cremona diagram in this way. We could keep solving the equilibrium of the truss with the node H, obtaining the internal forces in H-I and G-H, and so forth, but since the structure is symmetrical, we already know these internal forces, simply reading them on the other part and we know that here, we have compression ; here, tension in G-I; and then the internal force in G-J is equal to zero. I let you finish this drawing by yourself if you wish to. What we can see is that here, we have a N-shaped truss, with diagonals in tension. All the diagonals are in tension and that is due to this configuration in which the diagonals look towards the inside of the truss. Of course, this is only true if the loads act downwards. Let's note that when the diagonals are in tension, the posts are in compression. If you look at the equilibrium of the node C, for example, it is particularly clear, we have this tension which must be offset by this compression: it goes together. In this video, we can see a new truss structure, this time with a configuration in which the diagonals are orientated in the other way compared to what we have seen till now. I let you determine that this structure is statically determinate ; it is since it just has the same number of bars and nodes than the previous structure. Likewise, the loads being the same, the support reactions are identical and we can start the resolutiokn. We have a little problem here, because at the level of the node A, we have 3 bars ; one, two, three bars which meet thus we cannot proceed to the solving with the usual method. However, what we can observe is that the node B is again a particular node : this is a node with this kind of configuration. When we have this kind of configuration, the only way to have a polygon of forces which is closed, and then a structure in equilibrium, is that the internal forces in both these bars be equal to zero. So the internal force in these two bars, here, is equal to zero; and now, we can start to solve the structure beginning with node A. On this node A, the support reaction has a value of 15 N and is orientated upwards, then, turning counterclockwise from this force, we first meet an horizontal segment, A-C, and a segment orientated at 45Â° which is A-D Here is the equilibrium of the node A. We can see that the internal force in A-C pulls on the free-body, it is therefore a tensile internal force and that the internal force in A-B pushes on the free-body, it is thus a compressive internal force. The contribution of the node A to the Cremona diagram is this triangle. We now move to the node D : this node is subjected to a load of 10N, I draw it in a staggered way to be able to see it but it should be superimposed to the support reaction in A. A load of 10 N, then the internal force in A-D, in the other direction ; the internal force in C-D, vertical ; and finally the internal force in D-F, horizontal. The internal force in [CD] pulls on the free-body of the node D, it is therefore a tensile internal force. The internal force in D-F pushes on the free-body, it is a compressive internal force. Here is the contribution of the node D to the Cremona diagram. Let's now move to the node C : this node is subjected to the internal force in [CD], in the other direction, in A-C, in the other direction, then to the internal force in C-E -- I already know that this is going to be a tensile internal force, so I am going to draw it in tension, in a staggered way ; it should be superimposed. I am going to finish by the internal force in C-F, and here too, I already know that it is a compressive internal force, so I save some time by directly drawing it in blue. Here, this is the internal force in C-Fâ€¦ And here, this is the internal force in C-E. The contribution of the node C to the Cremona diagram is like this : we came here, we left in the other direction, we turned in this way. At this stage, we cannot solve node F yet, since there are still three bars in which the internal forces remain unknown here, however, we can solve node E, which is subjected to the internal force in C-E, in the other direction, and then -- I just redraw the tip of the arrow -- the internal force in E-G, which is identical, in the other direction. There cannot be any component of internal force inside E-F, because otherwise, it would be impossible for the vectorial sum to be zero with two vectors which are in this direction and one vector which is perpendicular to them. So, the internal force in E-F here must necessarily be equal to zero. The contribution of node E is then limited to, here, a small horizontal line. We can now move to the node F, which is first subjected to a new load of 10N, then to the internal force in D-F in the other direction, to the internal force in C-F in the other direction to zero in E-F, then to the internal force in F-G -- I know by symmetry that it is in compression so I directly continue -- to the internal force in F-G, in compression, and finally, to the internal force in F-H -- I also know by symmetry that it is also in compression. So here, I have the internal force in F-H, and here the internal force in F-G. The contribution of the node F has thus this shape of a house. As previously, we could keep solving the equilibrium moving to the node H, and we would obtain all the results, it would give us the following part of the Cremona diagram. However, since we know that our structure is symmetrical, we are not going to do it here. We have here an element in compression, and here two elements in tension. And finally here, two bars in which there is no internal force. We are thus here dealing with a N-shaped truss with diagonals in compression, and obviously, in this case, the postsâ€¦ â€¦ are in tension. We can see in this video that it is possible to release the bar which previously corresponded to G-J, without the equilibrium of the structure being impacted ; well, we had seen on this first truss that the internal force in G-J, and also the internal force in A-C, on the left, are equal to zero so we can obviously take off these chains -- at least, for this combination of load. This is due to the fact that we had a particular free-body here : when we have a free-body with a bar, and another bar in the other direction, and a force, from the support or not, it is obvious that here, we must have a compressive internal force which is equal to F to reach the equilibrium, and to get this same equilibrium, we must necessarily have an internal force which is equal to zero ; this is true for this node here, it is also true for this node there. We have another particular node, which we have also investigated, it is the node F. On this node F, we have the following situation : we have a bar which continues, and then a bar which is perpendicular to the first one, which are subjected to a certain force. For reasons of equilibrium, it is absolutely necessary that there is, here, a compressive internal force equal to F, -- this is what we have found in our solution-- while if we say the internal force in the chord -- for us, it was a compressive internal force -- if we say that we have an internal force N on the left, therefore the internal force on the right must be opposite. The Cremona diagram which is induced, if we start by the force F... we have the force F, then the internal force N, thereafter the internal force F in the vertical bar and we have again the internal force N, horizontal, acting on the right. We have also seen particular nodes, in the trusses with diagonals in compression; here, this is the same truss than previously, but I took off the bars which were not used in this construction and you can see that it still stands well. Indeed, here, we have the same configuration - except that it is on the top - which we have seen for the truss with diagonals in tension ; here we have two bars which have a random angle, by the way, they do not necessarily need to have an angle of 90 degrees, if we have this kind of configuration, we will also have, in case there is no load acting on the node, we will also have internal forces equal to zero in these bars. We had another particular node, here : it was the node E. This is a particular case of what we have just seen on the trusses with diagonals in tension, since we have here a bar that continues, in which there is an internal force, and we have here a vertical bar. We have here an horizontal internal force N, on the left and on the right, and it is necessary, to have equilibrium, that the internal force in this vertical E-F be equal to zero, otherwise, it is not possible to reach the equilibrium. In this lecture, we have seen how to solve the trusses with N-shaped diagonals, we have seen that it was sometimes necessary to start by a particular node, for which we can determine, for example, that certain internal forces are equal to zero, before being able to proceed to the usual solving. But it has always been possible to proceed with resolution. We have obtained the internal forces observing, in particular, that according to the orientation of the diagonals, they are systematically in tension or in compression, and that in addition, the posts are systematically in compression or in tension. We have observed a certain number of particular nodes, in which it is possible, simply observationally, to determine that the internal force is equal to zero or that it must be equal to another quantity, or to the internal force in the adjacent bar.